= 10-1/10! + 11-2/11! +.........+ 1000-991/1000!

=10/10! - 1/10! + 11/11! - 1/11! +....+ 1000/1000!-1/1000!

=1/9! - 1/10! + 1/10! - 1/11! +....+ 1/999! - 1/1000!

=1/9! - .1/1000! 

Ta thấy : 1/9! - 1/1000! < 1/9!

Cho mình hỏi bạn có phải là NGUYỄN THÚY HUYỀN _ LỚP 6B _ TRƯỜNG TRUNG HỌC CƠ SỞ VĨNH YÊN _ VĨNH PHÚC không ?

Ban lam co dung 100% hong

ban co chac chan la dung ko vay

Có: \(\frac{9}{10!}=\frac{9}{10!}\)

\(\frac{9}{11!}< \frac{10}{11!}=\frac{11-1}{11!}=\frac{11}{11!}-\frac{1}{11!}=\frac{1}{10!}-\frac{1}{11!}\)

\(\frac{9}{12!}< \frac{11}{12!}=\frac{12-1}{12!}=\frac{12}{12!}-\frac{1}{12!}=\frac{1}{11!}-\frac{1}{12!}\)

............

\(\frac{9}{1000!}< \frac{999}{1000!}=\frac{1000-1}{1000!}=\frac{1000}{1000!}-\frac{1}{1000!}=\frac{1}{999!}-\frac{1}{1000!}\)

\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{1}{1000!}< \frac{9}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{999!}-\frac{1}{1000!}\)

\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+...+\frac{1}{1000!}< \frac{10}{10!}-\frac{1}{1000!}=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\)

\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)

\(\Rightarrowđpcm\)

đặt tên là B

$\mathrm{B\; =}\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+.............+\frac{9}{1000!}$

Ta thấy :

$\frac{9}{10!}=\frac{10-1}{10!}=\frac{1}{9!}-\frac{1}{10!}$

$\frac{9}{11!}<\frac{11-1}{11!}=\frac{1}{10!}-\frac{1}{11!}$

$\frac{9}{1000!}<\frac{1000-1}{1000!}=\frac{1}{999!}-\frac{1}{1000!}$

$\Rightarrow B<\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+........+\frac{1}{999!}-\frac{1}{1000!}$

$B<\frac{1}{9!}-\frac{1}{1000!}$

$\Rightarrow B<\frac{1}{9!}\to đpcm$

Ta có:

\(\dfrac{9}{n!}\)< \(\dfrac{n-1}{n!}\) = \(\dfrac{1}{(n-1)!} - \dfrac{1}{n!}\) với n > 10 (n thuộc Z)

\(\Rightarrow\) \(\dfrac{9}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!} \)

= \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!}\)

\(\Rightarrow\) \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{1}{10!} - \dfrac{1}{11!} + \dfrac{1}{11!} - \dfrac{1}{12!} + ....\)

= \(\dfrac{1}{9!} - \dfrac{1}{1000!}\)

\(\Rightarrow \) \(\dfrac{9}{10!} + \dfrac{9}{11!} + ...+ \dfrac{9}{1000!} < \dfrac{1}{9!}\)

Chúc bn hc tốt.

\(\frac{9}{10!}+\frac{10}{11!}+...+\frac{999}{1000!}\)

= \(\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+...+\frac{1}{999!}-\frac{1}{100!}\)

= \(\frac{1}{9!}-\frac{1}{1000!}\)< \(\frac{1}{9!}\)( dpcm )

A = 9/10! + 9/11! + 9/12! + ...... + 9/1000! < 9/10! + 10/11! + 11/12! + ... + 999/1000! = B
9/10! = 1/9! - 1/10!
10/11! = 1/10! - 1/11!
...
999/1000! = 1/999! - 1/1000!
=> B = 1/9! - 1/1000! < 1/9!
=> A < 1/9! (dpcm)